Answer:
8.8 m and 52.5 m
Explanation:
The vertical component and horizontal component of water velocity leaving the hose are
![v_v = vsin(\alpha) = 25sin(53^0) = 25*0.8 = 19.97 m/s](https://tex.z-dn.net/?f=v_v%20%3D%20vsin%28%5Calpha%29%20%3D%2025sin%2853%5E0%29%20%3D%2025%2A0.8%20%3D%2019.97%20m%2Fs)
![v_h = vcos(\alpha) = 25cos(53^0) = 25*0.6 = 15 m/s](https://tex.z-dn.net/?f=v_h%20%3D%20vcos%28%5Calpha%29%20%3D%2025cos%2853%5E0%29%20%3D%2025%2A0.6%20%3D%2015%20m%2Fs)
Neglect air resistance, vertically speaking, gravitational acceleration g = -9.8m/s2 is the only thing that affects water motion. We can find the time t that it takes to reach the blaze 10m above ground level
![s = v_vt + gt^2/2](https://tex.z-dn.net/?f=%20s%20%3D%20v_vt%20%2B%20gt%5E2%2F2)
![10 = 19.97t - 9.8t^2/2](https://tex.z-dn.net/?f=10%20%3D%2019.97t%20-%209.8t%5E2%2F2)
![4.9t^2 - 19.97t + 10 = 0](https://tex.z-dn.net/?f=4.9t%5E2%20-%2019.97t%20%2B%2010%20%3D%200)
t = 3.49 or t = 0.58
We have 2 solutions for t, one is 0.58 when it first reach the blaze during the 1st shoot up, the other is 3.49s when it falls down
t is also the times it takes to travel across horizontally. We can use this to compute the horizontal distance between the fire-fighters and the building
![s_1 = v_ht_1 = 15*0.58 = 8.8 m](https://tex.z-dn.net/?f=s_1%20%3D%20v_ht_1%20%3D%2015%2A0.58%20%3D%208.8%20m)
![s_2 = v_ht_2 = 15*3.49 = 52.5m](https://tex.z-dn.net/?f=s_2%20%3D%20v_ht_2%20%3D%2015%2A3.49%20%3D%2052.5m)